Normed vector space

A normed vector space is a vector space equipped with a norm, whose induced metric is $d(x,y)=\Vert x-y\Vert$.

On continuous functions on a compact interval, evaluation at an interior point is not continuous in the integral norm. Narrow triangular peaks have fixed point value and integral tending to zero. Subtracting such a peak from the constant function one gives functions vanishing at the point and converging in the integral norm to one. Thus the evaluation kernel is not closed, although it is closed in the uniform norm.

In a normed vector space, this set consists of all points at a fixed positive norm distance $r$ from $x_0$. It also makes sense in an infinite-dimensional Hilbert space, unlike a specifically Euclidean sphere. In a real Hilbert space, the sphere in a normed vector space $\Vert u{\Vert }^2=2E_0>0$ has tangent variations $v$ satisfying $(u,v)=0$; this follows by differentiating the fixed squared norm. This is the geometric constraint used in fixed-energy initial-condition optimality.

The completion of a normed vector space $X$ is a Banach space containing an isometric dense copy of $X$. Construct it from Cauchy sequences modulo sequences whose difference tends to zero; define the norm by $\Vert [(x_n)]\Vert ={\lim }_n\Vert x_n\Vert$. Equivalently, the canonical evaluation map into the bidual space is an isometry by the Hahn-Banach theorem, and the closure of its image supplies a completion.

For $\lambda \ge 1$, a normed vector space $X$ is lambda-injective when every bounded linear operator from a subspace $Y$ of any normed space $Z$ into $X$ extends to $Z$ with the displayed norm bound. The space of bounded scalar functions on an index set is 1-injective: extend each coordinate functional by the Hahn-Banach theorem and reassemble it using the coordinate functional representation of an operator into bounded indexed functions.

A space is a lambda-injective normed space exactly when every linear isometry $J:X\to Z$ has a bounded left inverse as displayed. Necessity extends the inverse on $J(X)$. For sufficiency, embed $X$ in bounded scalar functions on an index set, extend the composed operator coordinatewise, and compose that extension with the assumed left inverse. The associated operator $JP$ is a projection onto $J(X)$.

The norm topology is the metric topology induced by $d(x,y)=\Vert x-y\Vert$.

A sequence converges in norm when $\Vert x_n-x\Vert \to 0$.

Every finite-dimensional vector space subspace of a normed vector space is closed. Coordinates in a finite basis identify it homeomorphically with a finite-dimensional scalar space; it is therefore complete, and every complete subspace of a metric space is closed.

An isometric isomorphism between normed spaces is a bijective linear map $T$ satisfying $\Vert Tx\Vert =\Vert x\Vert$ for every vector $x$.

A norm assigns a nonnegative length to each vector, is zero only at the zero vector, is absolutely homogeneous, and satisfies the triangle inequality.

A quasi-norm on a vector space is nonnegative, vanishes only at zero, is absolutely homogeneous, and satisfies a triangle inequality with one constant factor $C\ge 1$. For $0<p<1$, $(\int |f{|}^p{)}^{1/p}$ on an Lp space is a quasi-norm: $\Vert f+g{\Vert }_p^p\le \Vert f{\Vert }_p^p+\Vert g{\Vert }_p^p$ gives a quasi-triangle inequality. Measure-preserving composition preserves this quantity even though the Minkowski inequality is unavailable below exponent one.

The block mixed norm sums the ${\ell }^q$ norm of each block, for $q\ge 1$. Its dual norm is $\Vert p{\Vert }_{q^{\prime },\infty }={\max }_i\Vert p_i{\Vert }_{q^{\prime }}$, where $q^{\prime }=q/(q-1)$ for $q>1$ and $q^{\prime }=\infty$ for $q=1$. Apply Holder inequality blockwise and choose a norming block to attain equality. Absolute values are essential for signed components.

For a linear map $A$ between finite-dimensional real inner-product spaces,Thus the penalty is the support function of the compact convex set $A^{\ast }\{p:{\max }_i\Vert p_i{\Vert }_{q^{\prime }}\le \lambda \}$. The Moreau decomposition makes its proximal operator the identity minus the Euclidean projection onto a convex set.

Definiteness says that a norm vanishes only at the zero vector. The box norm on a nonempty finite Cartesian product is definite because its defining sum of squares includes equal-column terms, each of which is the square of an average of squares. Omitting repeated coordinates would destroy this proof and can produce a different quantity.

Absolute homogeneity is the norm axiom $\Vert av\Vert =|a|\Vert v\Vert$ for every scalar $a$ and vector $v$. It distinguishes a norm from a merely homogeneous expression of a different degree; for example the fourth root in the box norm converts a degree-four expression into a degree-one norm.

Suppose $B$ contains zero, is convex, and meets every line $\mathbb{R}v$, $v\ne 0$, in the finite symmetric open segment $\{tv:|t|<\lambda (v)\}$ with $\lambda (v)>0$. Then its Minkowski functional is a norm, with $p_B(v)=1/\lambda (v)$ and $B=\{x:p_B(x)<1\}$. Absolute homogeneity follows by rescaling the segment. If $s>p_B(x)$ and $t>p_B(y)$, convexity puts $(x+y)/(s+t)$ in $B$, giving $p_B(x+y)<s+t$; passing to the infima proves the triangle inequality. Conversely the open unit ball uniquely determines any norm by the displayed formula. No prior topology on the vector space is needed for these radial hypotheses.

On a uniform $d$-dimensional spatial grid, the discrete L2 norm weights the squared Euclidean norm by the cell volume. It approximates the continuous L2 norm and is induced by the discrete L2 inner product. A contraction bound in this norm with a mesh-independent constant proves stability of a numerical method for the spatially discretized evolution.

For $x\in {\mathbb{R}}^d$, the L1 norm is

For $0<c\le w_k<\infty$, interpret the nonnegative sum as an extended-valued functional on the l2 sequence space. Finiteness implies $u\in {\ell }^1$, but unbounded weights can make it infinite even there. It is proper and convex. Every finite partial sum is norm-continuous, so their supremum is sequentially lower semicontinuous and also weakly lower semicontinuous. This is useful as a sparsity penalty in variational regularization.

The coordinate conditions define a subgradient precisely when $p\in {\ell }^2$ and $J_w(u)<\infty$. They follow by varying one coordinate; summing their scalar supporting inequalities proves sufficiency. Since $w_k\ge c>0$, a subgradient can exist only at finitely supported $u$. At such a base point $v$, $D_{J_w}^p(u,v)={\sum }_k(w_k|u_k|-p_k{u}_k)$, allowing the value infinity. Merely having finite penalty does not guarantee a nonempty subdifferential in the ambient l2 sequence space.

Taxicab geometry equips Euclidean coordinate space with the distance induced by the L1 norm, so the distance between two points is the sum of their coordinatewise absolute differences.

The trace norm of a finite-dimensional operator $A$ is the sum of its singular values:For a Hermitian operator $A=A_{+}-A_{-}$, it equals $\mathrm{Tr}{A}_{+}+\mathrm{Tr}{A}_{-}$.

A Hermitian operator satisfies $\Vert X{\Vert }_1={\max }_{-I\le T\le I}\mathrm{Tr}(XT)$. Each diagonal entry of $T$ in an eigenbasis of $X$ lies between minus one and one, giving the upper bound ${\sum }_i|{\lambda }_i|$. The sign operator of $X$ attains it. The substitution $T=2E-I$ converts this formula to optimization over binary POVM effects and proves the Holevo–Helstrom theorem.

For any orthonormal basis and Hermitian operator, ${\sum }_i|⟨{\psi }_i|X|{\psi }_i⟩|\le \Vert X{\Vert }_1$. In the trace-norm variational principle for Hermitian operators, choose the diagonal $T$ whose entries are the signs of these diagonal entries. This bounds the total distinguishability visible in one fixed measurement basis.

The induced trace norm of a linear map $T$ between matrix spaces isIt measures the largest trace-norm amplification available without an auxiliary tensor factor.

The diamond norm is the stabilized induced trace normFor a map on $d$-dimensional inputs, an auxiliary space of dimension $d$ suffices. The tensor factor lets the norm detect how $T$ acts on one half of an entangled input.

The Euclidean norm of $x=(x_1,\dots ,x_n)\in {\mathbb{R}}^n$ is

For a real vector $v$, the square of its Euclidean norm is $\Vert v{\Vert }_2^2=v\cdot v={\sum }_j{v}_j^2$. Unlike the norm itself it is a quadratic function, so expanding sums directly exposes diagonal and cross terms. This is useful for the mean-square displacement of an isotropic planar random walk.

The Euclidean norm is self-dual:The upper bound is Cauchy-Schwarz inequality, and equality holds at $v=x/\Vert x{\Vert }_2$ when $x\ne 0$.

The closed Euclidean ball of radius $R$ about $a\in {\mathbb{R}}^n$ is $\{x:\Vert x-a{\Vert }_2\le R\}$; replacing $\le$ by $<$ gives the open ball.

The unit ball of a normed vector space is $\{x:\Vert x\Vert \le 1\}$ under the closed-ball convention.

The closed unit ball of a normed vector space is the set of vectors of norm at most one. In a continuous dual space it is compact in the weak-star topology by the Banach-Alaoglu theorem.

The open unit ball of a normed vector space $X$ is $B_X=\{x\in X:\Vert x\Vert <1\}$.

The supremum norm of a bounded real- or complex-valued function on a set $X$ isFor a continuous function on a compact space, the extreme value theorem makes this supremum a maximum.

On a nondegenerate interval, point evaluation of the derivative is unbounded for the supremum norm on smooth functions. The functions $f_N(x)=\sin (N(x-x_0))$ have supremum norm at most one but $f_N^{\prime }(x_0)=N$. Adding the supremum norm of the derivative makes this evaluation a bounded linear functional.

For a function with bounded derivatives through order $k$, the derivative supremum norm is $|u{|}_{k;U}={\sum }_{j=0}^k{\sup }_U|D^{j}u|$, with any equivalent finite-dimensional norm for each derivative tensor. This is a classical $C^k$ norm; it is different from the Lp norm with exponent $k$.

A surjective isometry $u:V\to W$ between normed vector spaces preserves every distance:

Every surjective isometry between real normed vector spaces is affine. In particular, if $u(0)=0$, then $u$ is real-linear.

For $v,w$ in a real normed space, start with the points at distance $\Vert v-w\Vert /2$ from both endpoints and repeatedly retain those within half the previous diameter of every retained point. Reflection about $(v+w)/2$ preserves every stage, the midpoint belongs to every stage, and the diameters decrease by a factor at least two. Their intersection is therefore exactly the midpoint.

Surjectivity in the Mazur-Ulam theorem is essential. The mappreserves all distances and fixes zero, but it is not linear and does not preserve every midpoint.

Two norms on one vector space are equivalent when constants $c,C>0$ satisfyfor every vector. Equivalent norms define the same convergent sequences, open sets, and continuous linear functionals.

Any two norms on a finite-dimensional real or complex vector space are equivalent. Relative to a fixed basis, continuity and positivity of a norm on the compact Euclidean unit sphere give uniform positive lower and finite upper bounds.

A Banach space is a normed vector space that is complete in its norm metric.

For a closed vector subspace $Y$ of a Banach space $X$, the quotient vector space $X/Y$ is complete with its quotient norm. Closedness makes the quotient seminorm a norm. Completeness follows by choosing representatives of a rapidly Cauchy subsequence whose successive differences have summable norms in $X$.

If a Banach space $X$ has separable dual and $q:X\to X/Y$ is a quotient by a closed subspace, every weakly null sequence in the quotient's closed unit ball has a subsequence with approximate weakly null lifts of norm at most $3$. Choose representatives of norm below $3/2$, pass to a common bidual weak-star limit, and subtract convex blocks whose quotient images tend to zero in norm. Convex-block cancellation of a weak-star limit gives weak nullity and the triangle inequality gives the bound $3$.

The quotient norm measures the smallest norm among representatives of a coset. Its infimum need not be attained. Nevertheless, for every $\epsilon >0$, an element $z\in X/Y$ has a representative $v\in X$ with $\Vert v\Vert <\Vert z\Vert +\epsilon$. The quotient map is contractive.

A convex block of a sequence in a Banach space is a sequence of convex combinations supported on successively disjoint finite intervals, with $p_n<q_n<p_{n+1}$, $a_i\ge 0$, and ${\sum }_{i=p_n}^{q_n}{a}_i=1$. Zero coefficients can fill gaps and ensure strict endpoint inequalities. By the Mazur theorem, every weakly null sequence has convex blocks tending to zero in norm.

If a bounded sequence $(y_n)$ in $X$ converges weak-star to $\varphi \in X^{\ast \ast }$ under the canonical embedding into the bidual, then any convex blocks $(u_n)$ have the same scalar limit on every $f\in X^{\ast }$. Thus $y_n-u_n$ is a weakly null sequence, even when $\varphi$ does not belong to the embedded space $X$.

The duality mapping takes a vector to its supporting dual functionals for half its squared norm. Equivalently $p\in \mathcal{J}(x)$ when $\Vert p{\Vert }_{\ast }=\Vert x\Vert$ and $⟨x,p⟩=\Vert x{\Vert }^2$. This follows from the dual-norm inequality and equality in the conjugate relation for squared norms. In a Hilbert space, the Riesz representation theorem identifies it with the identity; in a general Banach space it can be set-valued.

For an arbitrary set $\Gamma$ and scalar field $\mathbb{F}$, this is the space of bounded functions $u:\Gamma \to \mathbb{F}$ with the supremum norm $\Vert u\Vert ={\sup }_{\gamma }|u(\gamma )|$. Uniform limits of Cauchy sequences prove it is a Banach space. For $\Gamma =\mathbb{N}$ it is the l-infinity sequence space; for the empty index set it is the zero space, with norm zero.

A bounded linear operator $T:X\to {\ell }_{\infty }(\Gamma )$ is exactly a uniformly bounded family of bounded linear functionals $f_{\gamma }\in X^{\ast }$ through $(Tx)(\gamma )=f_{\gamma }(x)$. Evaluation shows $\Vert f_{\gamma }\Vert \le \Vert T\Vert$; conversely the family bound gives $\Vert Tx\Vert \le (\sup \Vert f_{\gamma }\Vert )\Vert x\Vert$, proving equality. Taking all functionals in the dual unit ball gives an isometric embedding by the Hahn-Banach theorem.

A Banach space is separable when it has a countable dense subset for its norm topology. It then has a countable base of norm balls and a countable norming family. Its continuous dual space need not be norm separable.

For $1\le p<\infty$, the space ${\ell }^p$ consists of scalar sequences $x=(x_n)$ satisfyingIt is a Banach space, and ${\ell }^q\subset {\ell }^p$ when $1\le q<p$.

The Banach space ${\ell }^1$ consists of absolutely summable scalar sequences, with norm $\Vert x{\Vert }_1={\sum }_n|x_n|$. Its continuous dual is ${\ell }^{\infty }$, under the pairing ${\sum }_n{x}_n{y}_n$. Finitely supported sequences are norm dense, which identifies a linear functional from its values on coordinate vectors. The space of sequences converging to zero inside this dual is a norming subspace of infinite codimension.

The space ${\ell }^2$ consists of scalar sequences $x=(x_n)$ satisfying ${\sum }_n|x_n{|}^2<\infty$. With inner product $⟨x,y⟩={\sum }_n{x}_n\overline{y_n}$, it is a Hilbert space.

The l2 sequence space is a uniformly convex Banach space. If $\Vert x{\Vert }_2=\Vert y{\Vert }_2=1$, the parallelogram law givesThus $\Vert x-y{\Vert }_2\ge \epsilon$ permits the modulus $\delta (\epsilon )=1-\sqrt{1-{\epsilon }^2/4}$.

A finitely supported sequence has only finitely many nonzero coordinates. Their vector space is commonly denoted $c_{00}$.

For $1\le p<\infty$, coordinate truncations converge in ${\ell }^p$, so $c_{00}$ is dense. It is not dense in ${\ell }^{\infty }$: every finitely supported sequence is at distance at least one from the constant-one sequence.

The space ${\ell }^{\infty }$ consists of bounded scalar sequences with norm $\Vert x{\Vert }_{\infty }={\sup }_n|x_n|$.

A Banach limit is a positive linear functional on the real l-infinity sequence space that extends the ordinary limit on the convergent sequence space, has operator norm one, and is invariant under the left shift on bounded sequences. The Hahn-Banach theorem constructs one by separating the constant sequence $1$ from the subspace $(I-S){\ell }^{\infty }$, whose distance from $1$ is one. The functional is generally not unique.

The left shift is a bounded linear operator of operator norm one on the l-infinity sequence space. For $y=x-Sx$, telescoping gives $N^{-1}{\sum }_{n=1}^N{y}_n=(x_1-x_{N+1})/N\to 0$. This identity is useful in constructing a Banach limit.

If $Y$ is a proper closed subspace of a normed space $X$ and $0<\alpha <1$, there is a unit vector $x\in X$ whose distance from $Y$ exceeds $\alpha$.

A normed vector space has compact closed unit ball if and only if it is finite-dimensional. In an infinite-dimensional space, repeated application of Riesz lemma produces unit vectors separated pairwise by a fixed positive distance, contradicting sequential compactness.

A Banach space $X$ is uniformly convex when, for every $\epsilon >0$, some $\delta >0$ satisfies

For $2\le p<\infty$, the Clarkson inequality isIt gives the uniform convexity of $L^p$.

If a family of bounded linear operators from a Banach space to a normed space is pointwise bounded, then their operator norms are uniformly bounded.

Let $(x_n)$ lie in a normed space $X$ and suppose ${\sum }_n|f(x_n)|<\infty$ for every $f\in X^{\ast }$. Then some $C$ satisfiesApply the uniform boundedness principle on the Banach space $X^{\ast }$ to the finite signed sums $f\mapsto {\sum }_n{a}_{n}f(x_n)$ with $|a_n|\le 1$.